Let $(a_{n,k}: n,k \ge 1)$ and $(b_{n,k}: n,k\ge 1)$ be nonnegative reals with the property that:
$\lim_{n\to \infty}\sum_{k=1}^\infty a_{n,k}=\lim_{n\to \infty}\sum_{k=1}^\infty b_{n,k}=1$;
for each $k\ge 1$, $\lim_{n\to \infty}a_{n,k}=\lim_{n\to \infty}b_{n,k}=0$.
(Implicitly, each limit and each infinite sum is well defined.)
Pick now a real bounded sequence $(x_k: k\ge 1)$ and assume that $(n_t: t\ge 1)$ is an increasing sequence of positive integers such that both limits $$ L_1:=\lim_{t\to \infty} \sum_{k=1}^\infty a_{n_t,k}x_k \quad \text{ and }\quad L_2:=\lim_{t\to \infty} \sum_{k=1}^\infty b_{n_t,k}x_k $$ exist.
Question. Is it true that $L_1=L_2$?
Attempt. Using the hypothesis that the sequence $x$ is bounded and the second property of the bullets, for each $\varepsilon>0$ and integer $K>0$, there exists a positive integer $n_0=n_0(\varepsilon, K)$ such that $$ \forall n\ge n_0, \quad \max\left\{ \left|\sum_{k< K}a_{n,k}x_k\right|, \left|\sum_{k< K}a_{n,k}x_k\right| \right\} \le \varepsilon $$ Moreover, the properties in the bullet imply that $$ \lim_{n\to \infty}\sum_{k\ge K} (a_{n,k}-b_{n,k})=0. $$
Assume $x\neq 0$, hence $\|x\|:=\sup_n|x_n|\neq 0$. Now, fix $\varepsilon>0$ and an integer $K>0$ and note that $$ \left|\sum_{k=1}^\infty a_{n_t,k}x_k-\sum_{k=1}^\infty b_{n_t,k}x_k\right| $$ is upper bounded, for all $t\ge n_0(\varepsilon,K)$ by $$ 2\varepsilon+\left|\sum_{k\ge K}(a_{n_t,k}-b_{n_t,k})x_k\right|\le 2\varepsilon+\|x\|\cdot \sum_{k\ge K}|a_{n_t,k}-b_{n_t,k}| $$ Is it true that the latter series is absolutely convergent to $0$ as $t\to \infty$? If the answer is affirmative, I am probably missing something obvious.
It is false. The counterexample:
For each $n\ge 1$, let $e_n$ be the sequence which is constantly $0$ except at the $n$th position on which it is $1$.
Set $A:=(a_{n,k}: n,k\ge 1)$ as it follows: the sequence of sequences $((a_{n,k}: k\ge 1): n\ge 1)$ is $(e_1,e_3,e_3,e_5,e_5,e_7,e_7, \ldots)$.
Similarly, set $B:=(b_{n,k}: n,k\ge 1)$ so that the sequence of sequences $((b_{n,k}: k\ge 1): n\ge 1)$ is $(e_2,e_2,e_4,e_4,e_6,e_6,\ldots)$.
Fix $x=(1,0,1,0,1,0,\ldots)$. Then $Ax$ is the constant sequence $1$ and $Bx$ is the constant sequence $0$.